Problem: $\dfrac{d}{dx}\left(\dfrac{\ln(x)}{\sqrt{x}}\right)=$
Answer: $\dfrac{\ln(x)}{\sqrt{x}}$ is the quotient of two, more basic, expressions: $\ln(x)$ and $\sqrt{x}$. Therefore, the derivative of the expression can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(\dfrac{\ln(x)}{\sqrt{x}}\right) \\\\ &=\dfrac{\dfrac{d}{dx}(\ln(x))\sqrt{x}-\ln(x)\dfrac{d}{dx}(\sqrt{x})}{(\sqrt{x})^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{\dfrac1x\cdot \sqrt{x}-\ln(x)\cdot \dfrac1{2\sqrt{x}}}{(\sqrt{x})^2}&&\gray{\text{Differentiate }\ln(x)\text{ and }\sqrt{x}} \\\\ &=\dfrac{\dfrac{1}{\sqrt{x}}-\dfrac{\ln(x)}{2\sqrt{x}}}{x}&&\gray{\text{Simplify}} \\\\ &={\dfrac{1}{x\sqrt{x}}-\dfrac{\ln(x)}{2x\sqrt{x}}}&&\gray{\text{Simplify}} \\\\ &=\dfrac{2-\ln(x)}{2x\sqrt{x}}&&\gray{\text{Common denominator}} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left(\dfrac{\ln(x)}{\sqrt{x}}\right)=\dfrac{2-\ln(x)}{2x\sqrt{x}}$ or any other equivalent form.